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The problem asks to evaluate the limit:

\[
\lim_{x \to 3} \left( \frac{x}{x - 3} \int_3^x \frac{\sin t}{t} \, dt \right)
\]

This expression involves a limit and a definite integral, which may present an indeterminate form when \(x \to 3\). Let’s break it down step-by-step:

### Step 1: Check the behavior of the integral as \(x \to 3\)

The integral is:

\[
I(x) = \int_3^x \frac{\sin t}{t} \, dt
\]

Since the limits of integration are from \(3\) to \(x\), if we substitute \(x = 3\), this integral becomes zero:

\[
I(3) = \int_3^3 \frac{\sin t}{t} \, dt = 0
\]

Hence, as \(x\) approaches \(3\), the integral goes to zero.

### Step 2: Analyze the limit

The limit expression is:

\[
\lim_{x \to 3} \left( \frac{x}{x - 3} I(x) \right)
\]

Substitute \(I(3) = 0\), and note that as \(x \to 3\), the denominator \(x - 3 \to 0\) and the integral tends to 0, leading to an indeterminate form \( \frac{0}{0} \). This suggests that we should apply **L'Hôpital's Rule** to resolve the indeterminate form.

### Step 3: Apply L'Hôpital's Rule

L'Hôpital's Rule can be applied to limits of the form \( \frac{0}{0} \) by differentiating the numerator and denominator with respect to \(x\).

Let the numerator be \(f(x) = x I(x)\) and the denominator be \(g(x) = x - 3\). Differentiating both:

- The derivative of the numerator, using the product rule, is:
  \[
  \frac{d}{dx}\left( x I(x) \right) = I(x) + x \cdot I'(x)
  \]
  where \(I'(x) = \frac{\sin x}{x}\).

- The derivative of the denominator is:
  \[
  \frac{d}{dx} (x - 3) = 1
  \]

Thus, applying L'Hôpital's Rule:

\[
\lim_{x \to 3} \frac{x I(x)}{x - 3} = \lim_{x \to 3} \frac{I(x) + x \cdot I'(x)}{1}
\]

### Step 4: Evaluate at \(x = 3\)

We know that \(I(3) = 0\), so we need to evaluate \(I'(3)\):

\[
I'(x) = \frac{\sin x}{x}, \quad \text{so} \quad I'(3) = \frac{\sin 3}{3}
\]

Thus, the limit becomes:

\[
\lim_{x \to 3} \left( 0 + 3 \cdot \frac{\sin 3}{3} \right) = \sin 3
\]

### Final Answer:

The value of the limit is:

\[
\boxed{\sin 3}
\]

Would you like further details on any part of this solution? Here are some related questions you might find useful:

1. How does L'Hôpital's Rule work in different indeterminate forms?
2. Can you explain more about the product rule for derivatives?
3. What are the conditions for applying L'Hôpital’s Rule?
4. How can we approximate integrals involving sine functions?
5. What other methods can be used to solve this type of limit problem?

**Tip:** When dealing with indeterminate forms like \( \frac{0}{0} \), always check whether L'Hôpital’s Rule or Taylor expansion can simplify the evaluation.

Evaluate the limit as x approaches 3: lim (x→3) (x / (x - 3)) ∫[3 to x] (sin(t) / t) dt

This example evaluates the limit as x approaches 3 of an expression involving a definite integral of sin(t)/t. The problem requires using L'Hôpital's Rule due to an indeterminate form and applies differentiation techniques. Suitable for college-level calculus.

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